这两道题其实实质上是一样的，其实本质上都是n个数的全排列
在复习dfs的时候其实单个模拟一下dfs的递归过程更好的理解代码是怎样实现的
还有就是终止递归的判断条件中无论条件怎样都要添加return

八皇后问题
https://www.acwing.com/problem/content/845/

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 20;

int n;
char g[N][N];
bool col[N], row[N], dg[N], updg[N];

void dfs(int u)
{
    if(u == n)
    {
        for(int i = 0; i < n; i ++) cout << g[i] << endl;
        puts("");

        return;
    }

    for(int i = 0; i < n; i ++)
        if(!col[i] && !dg[u + i] && !updg[u - i + n])
        {
            g[u][i] = 'Q';
            col[i] = dg[u + i] = updg[u - i + n] = true;
            dfs(u + 1);
            col[i] = dg[u + i] = updg[u - i + n] = false;
            g[u][i] = '.';
        }
}

int main()
{
    cin >> n;
    for(int i = 0; i < n; i ++)
        for(int j = 0; j < n; j ++)
            g[i][j] = '.';

        dfs(0);

        return 0;
}

排列数字问题
https://www.acwing.com/problem/content/844/

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;
const int N = 10;

int n;
int p[N];
bool st[N];

void dfs(int u)
{
    if(u == n)
    {
        for(int i = 0; i < n; i ++) printf("%d ", p[i]);
        puts("");

        return;
    }

    for(int i = 1; i <= n; i ++)
    {
        if(!st[i])
        {
            p[u] = i;
            st[i] = true;
            dfs(u + 1);
            st[i] = false;
        }
    }
}

int main()
{
    cin >> n;
    dfs(0);

    return 0;
}